The force acting on a body moving along the x | vedclass

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(B) For a body to be in equilibrium,the net force acting on it must be zero,i.e.,$F = 0$. From the graph,$F = 0$ at both $x = x_1$ and $x = x_2$.For s

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$x = x_2$

Vedclass · Answer

(B) For a body to be in equilibrium,the net force acting on it must be zero,i.e.,$F = 0$. From the graph,$F = 0$ at both $x = x_1$ and $x = x_2$.For stable equilibrium,the potential energy $U$ must be at a minimum,which implies that the force $F = -dU/dx$ must change from positive to negative as $x$ increases through the equilibrium point. Alternatively,in terms of the $F-x$ graph,the slope of the $F-x$ curve $(dF/dx)$ must be negative at the equilibrium point.At $x = x_1$,the slope of the $F-x$ curve is positive.At $x = x_2$,the slope of the $F-x$ curve is negative.Therefore,the body is in stable equilibrium at $x = x_2$.

The force acting on a body moving along the $x-$axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at

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